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Is the space $\psi$ (described in problem 5I of L. Gillman and M. Jerison, Rings of continuous functions, Springer Verlag, 1976) a F-Z-space (i.e, space with $cl(X-Z(f))$ is a zero set for every $f$ in $C(X)$)?

$\textbf{Clarification}$ A collection $\mathcal{A}$ of infinite subsets of $\mathbb{N}$ is said to be an almost disjoint family if $R\cap S$ is finite whenever $R,S\in\mathcal{A}$ and $R\neq S$.

Let $\mathcal{E}$ be a maximal almost disjoint family of subsets of $\mathbb{N}$. Let $D=(w_{E})_{E\in\mathcal{E}}$ be a set of new distinct points. Let $\psi=\mathbb{N}\cup D$. We shall give $\psi$ the topology as follows. The points in $\mathbb{N}$ are isolated. The neighborhood filter around a point $w_{E}$ is generated by sets of the form $w_{E}\cup R$ where $R\subseteq E$ and $E\setminus R$ is a finite set. In this space, $w_{E}\cup E$ is the one-point-compactification of the space $E$.

The problem in the book states that the space $\psi$ is completely regular, pseudocompact, but not countably compact. In particular, $\psi$ is not realcompact and $\psi$ is not normal. Furthermore, every subspace of $\psi$ is a $G_{\delta}$-set.

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    $\begingroup$ Can you please include the definitions of each thing that appears in this question? At least, you should take the time to define \psi $\endgroup$ – Jon Bannon Sep 2 '13 at 16:28
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    $\begingroup$ If you are trying to do Problem 5I, please give some details of where you are stuck $\endgroup$ – Yemon Choi Sep 2 '13 at 19:02
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    $\begingroup$ I checked problem 5I and I found nothing about F-Z-spaces in that problem. Also, I edited this question to include the definition of $\psi$ and information about this space to clear any confusion. I am voting to reopen this question. $\endgroup$ – Joseph Van Name Sep 3 '13 at 3:06
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    $\begingroup$ For a normal but not FZ-space, start with a Tychonoff plank and add a convergent sequence $\langle s_m^n:n<\omega\rangle$ to each point $[\omega_1, m]$ on the right-most vertical column. Then the set $S=\{s_m^n:n,m<\omega\}$ is a co-zero set whose closure is not $G_\delta$ hence not zero. $\endgroup$ – jonathanverner Sep 3 '13 at 10:15
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    $\begingroup$ Another way to construct examples of normal spaces which are even compact and connected but are not FZ-spaces is to look at the order topology. For instance, if one takes the long line and adds a copy of the unit interval $[0,1]$ at the end of the long line, one obtains a ordered space $([0,1)\times\omega_{1})\cup[0,1]$ which is compact and connected since it is a complete lattice. The interval $(0,1]$ on the very right is a cozero set, but the closure $[0,1]=\overline{(0,1]}$ is not a zero set in $([0,1)\times\omega_{1})\cup[0,1]$. $\endgroup$ – Joseph Van Name Sep 3 '13 at 14:53
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As I commented above, I think the answer is that a $\psi$-space need not be an FZ-space, and that a counterexample may be constructed from a Luzin gap. Here are the details, which did not fit into the comment.

We first construct the MAD family, which will give the counterexample. Start by splitting ${\mathbb N}$ into two infinite sets $X_0,X_1$. Then let ${\mathcal A}_0=\{A^0_\alpha:\alpha<\omega_1\}$ be an arbitrary AD family on $X_0$ and ${\mathcal A_1}=\{A^1_\alpha:\alpha<\omega_1\}$ a Luzin gap on $X_1$ (i.e. an AD family which, in particular, has the property that no uncountable pair of subfamilies can be separated; here ${\mathcal B},{\mathcal C}\subseteq{\mathcal A}_1$ can be separated if there is a set $S\subseteq X_1$ which almost-contains all sets from $B$ and is almost disjoint from all sets in $C$.).

We shall fuse these two families into a single new AD family which we will extend to a MAD family. Proceed as follows: for $\alpha<\omega_1$ let $A_\alpha=A^0_\alpha\cup A^1_{\alpha+1}$ and let $\{B_\alpha:\alpha<\omega_1\}$ be the enumeration of $\{A^1_\beta:\beta\in Lim(\omega_1)\}$. Let ${\mathcal A}$ be an arbitrary MAD family extending the family $\{A_\alpha,B_\alpha:\alpha<\omega_1\}$.

Proposition The space $\psi({\mathcal A})$ is not an FZ-space.

Note In the following we identify the points of $\psi({\mathcal A})={\mathbb N}\cup D$ which are in $D$ with the elements of the MAD family ${\mathcal A}$.

Proof. Define $f:\psi({\mathcal A})\to{\mathbb R}$ as follows: $f(n)=1/n$ for $n\in X_0$ and $f(x)=0$ otherwise. It is clear that $f$ is continuous. Let $G=\psi(\mathcal{A})\setminus f^{-1}(0) = X_0$. To show that $\psi({\mathcal A})$ is not an FZ-space it will be sufficient to show that $F=cl(G)$ is not a zero set. Notice first that for each $\alpha<\omega_1$ we have $A_\alpha\in F$ and $B_\alpha\not\in F$. Suppose now that $g:\psi({\mathcal A})\to{\mathbb R}$ is a continuous function such that $g^{-1}(0)=F$. Since $B_\alpha\not\in F$ for each $\alpha<\omega_1$, we can fix $n<\omega$ such that ${\mathcal B}=\{B_\alpha:g(B_\alpha)>1/n\}\subseteq{\mathcal A}_1$ is uncountable. Now let $S=\{k<\omega:g(k)>1/n\}$. Then $S$ almost contains all sets from ${\mathcal B}$ while, by continuity of $g$, being almost disjoint from every set in $\{A_\alpha:\alpha<\omega_1\}$ hence also from every set in ${\mathcal C}=\{A^1_{\alpha+1}:\alpha<\omega_1\}\subseteq{\mathcal A}_1$. This contradicts the fact that ${\mathcal A}_1$ is a Luzin gap, since $S$ separates the two uncountable subfamilies ${\mathcal B,C}\subseteq{\mathcal A}_1$. $\square$

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  • $\begingroup$ excuse me. I edit "FZ-space"----> "SZ-space" $\endgroup$ – Vahideh Bagheri Sep 7 '13 at 17:06

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